Proving function f(x) is a PDf given integral relationships

In summary, the problem is to prove that the given integral is equal to 1 and thus a probability density function, given that c is a maximum and that the integral of f(x) from 0 to pi^2 is equal to ln(pi^2 - e). The approach may involve expressing the integral in terms of a variable and solving for c, but further clarification on the relationship involving c is needed.
  • #1
Saracen Rue
150
10

Homework Statement


Prove that ##\int _0^a\left(\frac{f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\right)dx## is a probability density function (when ##a=\frac{1}{\pi ^2}##) given that ##\int _0^{\pi ^2}f\left(x\right)dx=\ln \left(\pi ^2-e\right)## and that ##c## is a maximum.

Homework Equations


##\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)##

The Attempt at a Solution


I'm honestly completely stuck with this question. I know that ##\int _0^a\left(\frac{f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\right)dx## can be expressed as ##\frac{1}{\ln \left(\pi ^2-e\right)+\pi ^2\sqrt{\cos \left(e-\pi ^2\right)}}\int _0^a\left(f\left((sin(ln(c))x\right)+\sqrt{\cos \left(e-\pi ^2\right)}\right)dx##, but I am unsure of how this helps me to prove that the integral is equal to 1 (thus proving it is a PDf). By specifying that ##c## is a maximum, the question insinuates that ##c## is a variable that can be expressed in terms of another variable which in turn can be derived and solved, however I am unsure of how to form such a relationship.
 
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  • #2
c is a maximum of what?

There are mismatched brackets related to f.
It looks like the problem depends on c, which is odd. What exactly do you know about c?
 

Related to Proving function f(x) is a PDf given integral relationships

1. What is a PDF (Probability Density Function)?

A PDF is a mathematical function that describes the probability distribution of a continuous random variable. This means that it shows all the possible values that a random variable can take and the likelihood of each value occurring.

2. How do you prove that a function f(x) is a PDF?

To prove that a function f(x) is a PDF, you need to show that it satisfies two conditions: it must be non-negative for all values of x and its integral over the entire range of x must equal to 1.

3. Can a function have multiple PDFs?

No, a function can only have one PDF. This is because a PDF is a specific representation of the probability distribution of a random variable, and a random variable can only have one probability distribution.

4. Can a PDF be used to determine the probability of a specific value?

No, a PDF cannot be used to determine the probability of a specific value. Instead, it gives the probability of a range of values. The probability of a specific value can be found by taking the limit of the PDF as the range approaches that value.

5. How is a PDF related to the cumulative distribution function (CDF)?

The CDF is the integral of the PDF, which means that it represents the cumulative probability of a random variable taking on values less than or equal to a certain value. In other words, the CDF is the area under the PDF curve, and the PDF is the derivative of the CDF.

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